Multiple ergodic averages with varying number of terms

2011-05-05T20:52:11Z

Hi. I've been stuck on the following question for some time.

Consider a sequence of functions $\left( f_n \right)$ from an ergodic space $\left( \mathsf{X}, \mathsf{S}, \mu \right)$ to $\left[ 0,1 \right]$ such that $f_{a+b} \leq f_a \mathsf{S}^a \left( f_b \right)$ for all integers $a, b \geq 0$, where, classically, $\mathsf{S} \left( g \right)$ is the map $x \longmapsto g \left( \mathsf{S} \left( x \right) \right)$.

Obviously $f_n \leq f_{n+1}$ so that the sequence $\left( f_n \right)$ decreases at each point to a function $f$. Under the hypothesis $\int_{\mathsf{X}} f_1 d\mu < 1$ I was able to prove that $f = 0$ almost everywhere. Here's how I dealt with the problem:

For fixed $n$ and integers $k$ and $\alpha$ such that $k \alpha \leq n$, the inequality \begin{equation} f_n \leq f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) \end{equation} holds, so that, taking averages, one gets \begin{equation} f_n \leq \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) \end{equation} Taking integrals, and using the dominated convergence theorem and the Furstenberg-Katznelson theorem on multiple ergodic averages, one gets \begin{equation} \int_{\mathsf{X}} f \leq lim_{n \longrightarrow \infty} \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} \int_{\mathsf{X}} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) = \left( \int_{\mathsf{X}} f_1 \right)^\alpha \end{equation} For all integers $\alpha$, which allows me to conclude.

Now the question is: does the series $\sum f_n$ converge? That seems plausible considering the seemingly exponential decreasing of $f_n$ to $f$, but trying to use the same techniques leads to multiple ergodic averages for a varying number of terms - making the integer $\alpha$ dependent on $n$ that is; is there any way to deal with those? All suggestions are welcome.

Answer by R W for Multiple ergodic averages with varying number of terms

2011-05-06T06:14:07Z

Unless I miss something, you are overcomplicating the story. In the first place, as it has already been pointed out, you don't really need submultiplicativity as $$ f_n(x) \le f_1(x) f_1(Sx) \dots f_1(S^{n-1} x) = F_n(x) \;. $$ Applying the usual ergodic theorem to the right-hand side one gets that a.e. $$ \log F_n(x)/n \to \int\log f_1(x)\;d\mu(x) < 0 \;, $$ whence the claim.

Multiple ergodic averages with varying number of terms - MathOverflow

Multiple ergodic averages with varying number of terms

Answer by R W for Multiple ergodic averages with varying number of terms